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I already solved for the homogeneous one, but I'm still looking for the particular solution of the differential equation:

$$y''+3y'+2y=\exp(\mathrm{e}^x)$$

The homogeneous solutions of this system are $\mathrm{e}^{-x}$ and $\mathrm{e}^{-2x}$. I've tried the substitution

$$y_p(x)=v(x)\mathrm{e}^{-2x},$$ which yielded the differential equation:

$$v''+v'=\exp(\mathrm{e}^x+2x),$$

After that I reduced the order by $p=v'$ which gives:

$$p'+p=\exp(\mathrm{e}^x+2x).$$

Now this equation still seems hard to solve. I was wondering if there were easier/better substitutions to make ?

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    $\begingroup$ I have the feeling that it should be $v''-v'=\exp(e^x+2x)$ $\endgroup$ – Claude Leibovici Oct 10 '14 at 9:58
  • $\begingroup$ Have you learned the Wronskian? Also, that last DE can be solved easily with an integrating factor. $\endgroup$ – UserX Oct 10 '14 at 10:00
  • $\begingroup$ @UserX, the Wronskian is known to me :). $\endgroup$ – Nick Oct 10 '14 at 11:01
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We first observe that $$ y=\frac{1}{(n+1)(n+2)}\mathrm{e}^{nx}, $$ is a particular solution of $$ y''+3y'+2y=\mathrm{e}^{nx}. $$ Hence $$ \sum_{n=0}^\infty \frac{1}{(n+1)(n+2)n!}\mathrm{e}^{nx}=\mathrm{e}^{-2x}\left(\exp(\mathrm{e}^x)-1-\mathrm{e}^x\right) $$ is a particular solution of $$ y''+3y'+2y=\sum_{n=0}^\infty\frac{1}{n!}\mathrm{e}^{nx}=\exp(\mathrm{e}^x). $$

Note. However, if one wants to be rigorous, a verification would be the right thing to do.

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  • $\begingroup$ this includes the complentary solution, because using my method above I have found a partial amount of your answer. However, when added with my complementary solution, I get the same answer as you do. My $c_1$ and $c_2$ are arbitrary constants. Meaning, you have found the general solution, not the particular solution. $\endgroup$ – Varun Iyer Oct 10 '14 at 10:38
  • $\begingroup$ @YiorgosS.Smyrlis, that's a very nice trick :) $\endgroup$ – Nick Oct 10 '14 at 11:04
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This solution might be more time-comsuming, but it can work.

We can use a method called variation of parameters.

If we take our characteristic polynomial:

$$y_c(t) = r^2+3r+2 = (r+2)(r+1)$$

Now we have that our complementary solution is:

$$y_c(t) = c_1e^{-2t} + c_2e^{-t}$$

Now our solution to this equation is $e^{-2t}$ and $e^{-t}$

We can use this to find our Wronskian

$$W = \begin{vmatrix}e^{-2t} &e^{-t} \\ -2e^{-2t} & -e^{-t} \end{vmatrix} = -e^{-3t}+2e^{-3t} = e^{-3t}$$

Now we can find our particular solution.

$$y_p(t) = -e^{-2t}\int\frac{e^{-t}e^{e^t}}{e^{-3t}}dt + e^{-t}\int\frac{e^{-2t}e^{e^t}}{e^{-3t}}dt$$

Solving this will get you the particular solution.

EDIT

Solving for the particular we get

$$y_p(t) = -e^{-2t}\left(e^{e^t}(e^{t}-1)\right) + e^{-t}e^{e^t}$$ $$y_p(t) = -e^{-2t}(e^{e^t}e^{t}-e^{e^t}) + e^{-t}e^{e^t} = e^{-2t}e^{e^t}$$

If you need the general solution, add both the complementary and the particular solution together.

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I believe Claude's comment is correct an that differential equation is not pretty. So I'll attempt a different approach, attempting to solve for the general solution.

Since the characteristic polynomial is $(s+2)(s+1)$, it appears integrating factors of $e^x$ or $e^{2x}$ will work. $e^x$ will result in easier integration. So we let

$$z=y'+2y$$ $$y''+3y'+2y=(y'+2y)'+y'+2y=z'+z=e^{e^x}$$ $$e^xz'+e^xz=(e^xz)'=e^xe^{e^x}$$ $$e^xz=e^{e^x}+k_1$$ $$e^x(y'+2y)=e^{e^x}+k_1$$ $$e^{2x}(y'+2y)=(e^{2x}y)'=e^xe^{e^x}+k_1e^x$$ $$e^{2x}y=e^{e^x}+k_1e^x+k_2$$ $$y=e^{e^x-2x}+k_1e^{-x}+k_2e^{-2x}$$

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